$f(x+f(y))=f(x)+f(f(y))+3xy^6+3x^2f(y)$

Question

Find all functions $f : \mathbb{R} \mapsto \mathbb{R}$ that check $f(x+f(y))=f(x)+f(f(y))+3xy^6+3x^2f(y)$

for all $x,y \in \mathbb{R}$ It's a very difficult functional equation, I tried many substitution

Can someone help me, please

Answer 1

This took me way too long :)

To shorten things up, I looked for an $n$ where $f(n)=n$ (as $f(0)=0$, there must be at least one). I did what you did – substituting $x$ and $y$ to a bunch of values. Here's the ones that yielded some results:

1. $x=-f(y)$
   $f(0)=f(-f(y))+f(f(y))-3y^6f(y)+3[f(y)]^3$ (1)
    

2. $x=f(y)$
   $f(2f(y))=2f(f(y))+3y^6f(y)+3[f(y)]^3$
   $0=2f(f(y))-f(2f(y))+3y^6f(y)+3[f(y)]^3$ (2)
    
   When $f(y)=y$:
      $f(2y)=2y+3y^7+3y^3$ (3)

As (1) and (2) both equal 0:
 
$f(-f(y))+f(f(y))-3y^6f(y)+3[f(y)]^3=2f(f(y))-f(2f(y))+3y^6f(y)+3[f(y)]^3$

$f(-f(y))=f(f(y))-f(2f(y))+6y^6f(y)$

Furthermore, since I'm looking for $y=f(y)$:
    $f(-y)=y-f(2y)+6y^7$ (for $y=f(y)$)
 
    We can then insert (3) into that:
    $f(-y)=y-(2y+3y^7+3y^3)+6y^7$
    $f(-y)=-y-3y^3+3y^7$
 
    I multiplied everything by $-1$, and with some luck got:
    $f(y)=y+3y^3-3y^7$
    $y=y+3y^3-3y^7$ (as $f(y)=y$)
    $0=3y^3(1-y^4)$
    $y^4=1\implies y=-1\bigcup y=1$

...and that worked! Putting those to the functions work correctly, including "larger" numbers (further away from zero) than $-1$ and $1$. $f(1)=1$, and $f(-1)=-1$.

To get what $f$ actually is, I substituted $y$ with $-1$:     $f(x+f(-1))=f(x)+f(f(-1))+3x(-1)^6+3x^2f(-1)$
    $f(x-1)=f(x)-1+3x-3x^2$
    $f(x)=f(x-1)+3x^2-3x+1$

Or, as @Gonçalo stated in their comment, simply $f(x)=x^3$ (I'm guessing $f$ can't be anything else).

Maybe the part where I multiplied everything by $-1$ doesn't always work. I know that it should've been $-f(-y)$ instead of $f(y)$ if I were doing it properly, so I suppose I had some luck.

Also, apologies if the formatting looks ugly. I tried to make it look more bearable, haha

Answer 2

You have already obtained the following identities: $$ \begin{align} &f(x+f(y))=f(x)+f(f(y))+3xy^6+3x^2f(y)\quad\forall x,y\in\mathbb R,\tag{1}\\ &f(0)=0,\tag{2}\\ &x^6=f(x)^2+cf(x)\quad\forall x\in\mathbb R.\tag{3} \end{align} $$ From $(3)$ we obtain $$ f(x)=\frac{-c+\sigma(x)\sqrt{c^2+4x^6}}{2}\quad\forall x\in\mathbb R\tag{4} $$ for some function $\sigma:\mathbb R\to\{-1,1\}$. It follows that $f$ is not bounded.

Pick an arbitrary real number $y$. Let $-x=z=f(y)$. Then by $(1)$, we get $f(0)=f(-z)+f(z)-3zy^6+3z^3$. In turn, by $(2)$ and $(3)$, we get $0=f(-z)+f(z)-3z(z^2+cz)+3z^3$, i.e., $$ f(z)+f(-z)=3cz^2\quad\text{whenever $z$ is in the range of $f$}.\tag{5} $$ Since $f$ is unbounded, we may pick some $z$ from the range of $f$ such that $|z|$ is sufficiently large. If $\sigma(z)=\sigma(-z)$, then $(4)$ gives $f(z)+f(-z)=-c+\sigma(z)\sqrt{c^2+4z^6}$ and hence $(5)$ gives $-c+\sigma(z)\sqrt{c^2+4z^6}=3cz^2$, which is impossible because $|z|$ is large. So, we must have $\sigma(z)=-\sigma(-z)$. But then $(4)$ and $(5)$ give $-c=3cz^2$. Hence $c=0$ and $(4)$ becomes $f(x)=\sigma(x)|x|^3$ for all $x\in\mathbb R$. Define $s:\mathbb R\to\{-1,1\}$ by $s(0)=1$ and $s(x)=\sigma(x)\operatorname{sign}(x)$ for all $x\ne0$. Then $$ f(x)=s(x)x^3\quad\forall x\in\mathbb R.\tag{6} $$ Now, for any $a\in\mathbb R$, let $y=a^{1/3}$. By $(6)$, $f(y)=s(y)y^3$. So, if we put $x=-s(y)y^3$ into $(1)$, we obtain $$ f(-s(y)y^3+s(y)y^3)=f(-s(y)y^3)+f(s(y)y^3)+3(-s(y)y^3)y^6+3(-s(y)y^3)^2(s(y)y^3), $$ i.e., $0=f(-s(y)y^3)+f(s(y)y^3)$. Since $s(y)=\pm1$, one of $-s(y)y^3$ or $s(y)y^3$ must be equal to $a$. Therefore $$ f(a)+f(-a)=0\quad\forall a\in\mathbb R $$ and from $(2)$ and $(6)$ we conclude that $s$ is an even function on $\mathbb R\setminus\{0\}$ and $f$ is an odd function on the real line.

We now show that $s(x)=s(1)$ for all $x>0$. Suppose the contrary. Then $f(x)=-s_1x^3$ for some $x>0$, where $s_1=f(1)=s(1)\in\{-1,1\}$. Since $f$ is odd, $f(f(1))=1$. Let $s_2=s(x+s_1)$. Put $y=1$ into $(1)$, we obtain $$ s_2(x+s_1)^3 = -s_1x^3 + 1 + 3x + 3s_1x^2. $$ Expand the LHS and rearrange terms, we obtain $$ (s_1+s_2)x^3 + 3s_1(s_2-1)x^2 + 3(s_2-1)x + (s_1s_2-1) = 0, $$ which is reduced to $$ \begin{cases} 2x^3 = 0&\text{when } (s_1,s_2)=(1,1),\\ -6x^2 -6x -2 = 0&\text{when } (s_1,s_2)=(1,-1),\\ -2 = 0&\text{when } (s_1,s_2)=(-1,1),\\ -2x^3 + 6x^2 -6x = 0&\text{when } (s_1,s_2)=(-1,-1).\\ \end{cases} $$ But then we arrive at a contradiction because none of these equations has a positive real root $x$.

Hence we must have $s(x)=s(1)$ for all $x>0$. Since $f$ is an odd function, we infer from $(6)$ that either $f=(x\mapsto x^3)$ or $f=(x\mapsto-x^3)$ on the whole real line. It is easy to check that among these two choices, the only solution is $f=(x\mapsto x^3)$.